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kondaur
15 days ago
10

Oceanographers use submerged sonar systems, towed by a cable from a ship, to map the ocean floor. In addition to their downward

weight, there are buoyant forces and forces from the flowing water that allow them to travel in a horizontal path. One such submersible has a cross section area of 1.3m2 , a drag coeffecient of 1.2, and when towed at 4.3 m/s, the tow cable makes an angle of 30 degrees with the horizontal. What is the tension in the cable? Take the water density to be 1000 kg / m3
Physics
1 answer:
Yuliya22 [3.3K]15 days ago
7 0
The tension exerted in the cable amounts to T = 16653.32 N. Parameters: Cross section area A = 1.3 m² Drag coefficient CD = 1.2 Velocity V = 4.3 m/s The angle formed by the cable with the horizontal is 30 degrees. Density defined as follows: The drag force FD is determined by the equation: FD = (1/2) * ρ * V² * A * CD Calculating the drag force yields 14422.2 N acting opposite to motion. Given the cable's angle of 30 degrees with horizontal, the horizontal component contributes to the drag force calculation: T * cos(30) = F_D Thus, T = 16653.32 N.
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A 0.050-kg lump of clay moving horizontally at 12 m/s strikes and sticks to a stationary 0.15-kg cart that can move on a frictio
Ostrovityanka [3204]

Answer:

Explanation:

According to the parameters provided,

mass of the clay lump, m₁ = 0.05 kg

initial velocity of the lump, u₁ = 12 m/s

mass of the cart, m₂ = 0.15 kg

initial speed of the cart, u₂ = 0

As the clay adheres to the cart, we have an inelastic collision scenario. Let v represent the combined speed of both the cart and lump post-collision. Given that momentum is conserved, we have:

m_1u_1+m_2u_2=(m_1+m_2)v

v=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}

v=\dfrac{0.05\ kg\times 12\ m/s+0}{0.05\ kg+0.15\ kg}

The resultant speed is v = 3 m/s.

Thus, the final speed of both cart and lump following the collision is 3 m/s. This concludes the solution.

3 0
1 month ago
An object moving with a speed v0 collides head-on with a second object initially at rest. A student assumes the collision is ela
serg [3582]
The answer is B. Since the first collision is elastic, both momentum and kinetic energy can be conserved within the system. The coefficient of restitution for an elastic collision is one, and it is often referred to as a perfectly elastic collision. Conversely, in a perfectly inelastic collision, kinetic energy is lost as it transforms into another form, such as internal energy. While momentum remains conserved in an inelastic collision, kinetic energy is not.
6 0
11 days ago
The 8 kg block is then released and accelerates to the right, toward the 2 kg block. The surface is rough and the coefficient of
kicyunya [3294]
3.258 m/s Explanation: The spring constant is assumed to be 263 N/m and the displacement of the spring is also assumed to be 0.7 m; the coefficient of friction between blocks is 0.4. The energy stored in the spring is described by . Given the conservation of energy in the system, the speed of the 8 kg block just prior to collision is 3.258 m/s.
7 0
1 month ago
a. For a spring-mass oscillator, if you double the mass but keep the stiffness the same, by what numerical factor does the perio
kicyunya [3294]

Answer:

a) factor b=\sqrt{2}

b) factor b=\frac{1}{2}

c) factor b=1

d) factor b=1

Explanation:

For an oscillating spring-mass system, the time period is expressed as:

T=\frac{1}{f}

T={2\pi} \sqrt{\frac{m}{k} }

where:

f= represents the frequency of oscillation

m= signifies the mass linked to the spring

k= is the spring's stiffness constant

a) If the mass is doubled:

  • New mass, m'=2m

Thus, the new time period:

T'=2\pi\sqrt{\frac{m'}{k} }

T'=2\pi\sqrt{\frac{2m}{k} }

T'=\sqrt{2}\times 2\pi\sqrt{\frac{m}{k} } }

T'=\sqrt{2} \times T

this leads to factor b=\sqrt{2} as per the question.

b) When the stiffness constant is quadrupled, holding other factors constant:

New stiffness constant, k'=4k

Thus, the new time period:

T'=2\pi\sqrt{\frac{m}{k'} }\\\\T'=2\pi\sqrt{\frac{m}{4k} }\\\\T'=\frac{1}{2} \times 2\pi\sqrt{\frac{m}{k} } }\\\\T'=\frac{1}{2} \times T

this results in factor b=\frac{1}{2} as required.

c) When both mass and stiffness constant are quadrupled:

New stiffness, k'=4k

New mass, m'=4m

Thus, the new time period:

T'=2\pi\sqrt{\frac{m'}{k'} }\\\\T'=2\pi\sqrt{\frac{4m}{4k} }\\\\T'=1 \times 2\pi\sqrt{\frac{m}{k} } }\\\\T'=1 \times T

which leads to factor b=1 as stated in the question.

d) If amplitude is quadrupled, the time period remains unaffected because T does not depend on amplitude as demonstrated by the equation.

Thus, factor b=1

7 0
1 month ago
The force shown in the attached figure is the net eastward force acting on a ball. The force starts rising at t = 0.012 s, falls
ValentinkaMS [3465]
Impulse can be expressed as the integral of F(t) dt from 0.012 s to 0.062 s

Since the function F(t) is unknown, an estimation method will be applied.

The integral represents the area under the curve.

The task suggests approximating this area as a triangle.

For this triangle, the base measures: 0.062 s - 0.012 s = 0.050 s.

The height corresponds to the maximum force of 35 N.

Consequently, the area computes to [1/2] * (0.05 s) * (35 N) = 0.875 N*s.

Final result: 0.875 N*s
6 0
1 month ago
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